Find the sum of the geometric series $1 - 0.99 + 0.99^2 - 0.99^3 +... - 0.99^{79}$ Choose 1 answer: Choose 1 answer: (Choice A) A $-0.45$ (Choice B) B $0.28$ (Choice C) C $0.73$ (Choice D) D $55.24$
Getting started We're dealing with a geometric series because each term is multiplied by $-0.99$ to get the next term. We need a formula to compute the sum of the terms. Formula for geometric series The sum $S_n$ of a finite geometric series is $S_n = \dfrac{a_1(1-r^n)}{1-r}$ where $a_1$ is the first term, $r$ is the common ratio, and $n$ is the number of terms. What do we need to use the formula? The first term $(a_1 = {1})$ is given in the question. The number of terms $n$ is ${80}$ because there are ${80}$ numbers from $0$ to $79$. [Where do the 0 and 79 come from?] The common ratio $r$ is ${-0.99}$ because each term is multiplied by ${-0.99}$ to get the next term. [How did we find the common ratio r?] Find the sum $(S_n)$ of the series $\begin{aligned} S_n &= \dfrac{a_1(1-r^n)}{1-r} \\\\ S_{{80}}&=\dfrac{{1}(1-({-0.99})^{{80}})}{1-({-0.99})} \\\\ S_{{{80}}} &\approx 0.28 \end{aligned}$ The answer $0.28$